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Let's consider the manifold $S^1$

It is well known that we need two charts to cover this manifold.

Nonetheless, we can cover the full space using a single coordinate $\theta$ which is just the angle from the center.

Now, is this a general feature? I mean, is it always possible to have in every manifold a single coordinate set that cover points that are in different charts, just as in $S^1$?

PhoenixPerson
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  • Coordinates are assumed to be unique on a chart, whence we want the corresponding map from an Euclidean domain to the manifold to be injective. It seems to me that you are trying ask which manifolds are continuous (or smooth) images of Euclidean domains. – Joonas Ilmavirta Aug 17 '14 at 16:15
  • It's not entirely clear to me what you're asking. What restrictions do you want on the coordinate. A thought I have is maybe: given a connect manifold $M$ is there a continuous (possibly some smoothness conditions on interior) bijection $S \to M$ for some connected subset $S \subset \mathbb{R}^m$. We might also want to mandate that $S$ be contained in the closure of its interior. – jxnh Aug 17 '14 at 16:16
  • what prompted me to ask this is the following. It has been told to me that manifolds are spaces that look like euclidean space just around every point and that we have to consider many charts to cover it entirely. After being told this they tell me that $S^1$ needs at least two charts but immeadiately after they use only ONE set of coordinate to parameterize it. Same stuff with $S^2$. I DO realize that in these cases the unique set of coordinates that are used just sew in a smooth way different patches so I thought that maybe this is a general case. – PhoenixPerson Aug 17 '14 at 18:29
  • What bothers me is that they tell you "you have to separate the manifold in some euclidean space look alike patches" and then everyone uses coordinates such as $\theta$ even though it extends itself over two patches – PhoenixPerson Aug 17 '14 at 18:33
  • @silvrfück: I know an example in R^2 which is covered by a single chart. – Mikasa Aug 17 '14 at 19:03
  • @BabakS. me too $S^1\times{}\mathbb{R}$; the key of my question is why and when people can use coordinates that cover may charts – PhoenixPerson Aug 17 '14 at 21:34
  • Could you explain more about how you've seen one set of coordinates to cover the $2$-sphere? There's no obvious analogy with your example of using an angle parameter $\theta$ on the circle. – Andrew D. Hwang Aug 17 '14 at 21:44
  • with $\phi$ and $\theta$ (please don't misunderstand this, I understand that two charts are needed to cover $S^2$ in a way that we get open subsets of $R^2$). Nonetheless it is true that with $\phi$ and $\theta$ we cover the full sphere. No one ever uses two coordinate systems belonging to two charts. Everyone always use $\theta$ and $\phi$ I wanted to know if it is always or when it is possible to do such a thing and use coordinates that overlap different charts – PhoenixPerson Aug 17 '14 at 21:49
  • If it matters, $\phi$ and $\theta$ aren't coordinates (in any reasonable sense) at the poles. Geographically, longitude fails to be well-defined at the poles in a more degenerate way than longitude fails to be well-defined elsewhere on the sphere. It does turn out that for every compact manifold $M$, there exists a coordinate chart covering a dense open subset of $M$. However, I don't know of a "low-tech" proof, e.g., one using only basic definitions. – Andrew D. Hwang Aug 17 '14 at 22:34
  • Circle without the north pole is homeomorphic to the real line, see for example this question – rych Sep 11 '14 at 08:42

3 Answers3

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No, your "single coordinate" is only a smooth (continuous) function on $S^1-(1,0)$.

Ted Shifrin
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By definition a coordinate system is a collection of charts $\{U_i,\phi_i\}_i$ where $U_i$ are open sets in your manifold that cover your whole manifold and $\phi_i:U_i \rightarrow \mathbb{R}^n$ are homeomorphisms and the transition functions $\phi^i \circ \phi^{-j}$ are smooth where ever defined.

In this case your only chart is $(S^1,\theta)$ and but if $\theta$ was a homeomorphism this means that $\theta(S^1)$ is an open interval in the real line (since $S^1$ is connected). Therefore you can build a homeomorphism between $S^1$ and $\mathbb{R}$ which is a contradiction. You can also apply this arguement to any $S^n$ to say they won't admit single charts.

The easiest examples of manifolds admitting single charts are obviously open sets of $\mathbb{R}^n$.

Sina
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Perhaps you're "really" asking whether or not every $n$-manifold $M$ has $\mathbf{R}^{n}$ as a covering space, i.e., whether there's a single (smooth) map $\pi:\mathbf{R}^{n} \to M$ whose restrictions to suitably small subsets define coordinate charts on $M$.

If so, the answer is "no"; even among spheres, only $S^{1}$ has this property.

Andrew D. Hwang
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